text:
What does the altimeter indicate if it's at a certain elevation
The purpose of the altimeter is to show the pilot the altitude and the pressure above a fixed level. It uses barometric pressure and is 29.92Hg in ISA conditions.
What I dont understand is, how can this be extrapolated to be used in a situation where there is an elevation. For example, say the airplane is in a static position in an airfield at code EGNX, 52°49′52″N 001°19′40″W that has a 10 foot elevation. What would the altimeter indicate in such a situation.
Edit: Added a starting position

An altimeter is just an adjustable barometer with extra hands.
There is a barometric correction you dial in so that the altitude shown is always feet above sea level.  The correction is from the current barometric pressure reference you get with the "altimeter setting" or the weather report.
You dial in the altimeter setting, the current barometric pressure reference for your location, and shows your true height above sea level.  In your example, if the altimeter setting is correct, it will show 10 ft above 0 ft.
Or, you can work it the other way.  If you know the elevation you are at, you can dial the altimeter so its altitude indication matches your known elevation.  Voila!; the current barometric pressure is shown in the setting window on the instrument's face.

Could this also work if I'm trying to use DMS to and an elevation to figure out the altimeter reading?

The altimeter, when properly set, shows the height above sea level.
When you're on the ground at Denver International Airport, at 5,434 feet above sea level, your altimeter will show 5,434 on the ground.
When you're on the ground at San Francisco, nearly at sea level, your altimeter will show 13 feet.

Specific notes:
"when properly set" means that it has to be set according to the local barometer setting.  If the pilot has set the incorrect barometer setting, it will show the wrong altitude.

'when properly set' - the altimeter will only show height above sea-level when set to the area QNH. If properly set to the QFE it will show height above the airfield. If properly set to 1013.2hPa (29.92"" Hg) it will act as a reference for flight level.

If the initial airfield location is given in DMS form and then the elevation of 10 feet is mentioned. How would the result of the altimeter be calculated.

The altimeter doesn't know anything about location, whatever form you give it in. It's not a GPS receiver.

Okay in that case, I am really confused as to how to go about solving this question. I will enter the inputs that I am given; Departure: code EGAC, Arrival: code EGNX, the planned flight time is 40 minutes, the total flight distance is approximately 200NM, the cruise flight level is 29,000 feet, the cruise speed is at M0.73. The question states that the flight will take off from the airport code EGAC, and the airport elevation is 150 feet. What does the altimeter currently indicate? @CatchAsCatchCan

@SillySquishy So, this isn't a question about altimeters, but a question about procedure. Strictly speaking, you can't answer the question because you haven't been given the area QNH or the actual altimeter setting. However, as a matter of procedure, ATC will give you the area QNH prior to departure.With that set on your altimeter you'll get height above sea level, which, for your question is 150ft.

EGAC is at 15ft MSL (https://skyvector.com/airport/EGAC/Belfast-City-Airport)

I have access to the mark scheme which has the answer at -240 ft, however there isn't any working out as to how this answer was derived. & yes @abelenky i meant 15 feet elevation not 150 feet, sorry for the mistake.

@SillySquishy Well, there's no way to get to that answer from the information you've given. There must be some context elsewhere.

Well thanks for your effort, I will try to hunt down the person who wrote this question and see if i can get a worked solution from them. Thanks again :)


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When was a multi-engine rating first required?
For the United States, does anyone know how this progressed over time / when?
Was it always a rating?  Was it an endorsement at some time?  Called something else?
Any history out there?
Thanks

Please specify whether you mean in the US, or somewhere else, or anywhere in the world-- thanks -- and should be in the body of the question (or title) as well as just the tags-- thanks

@quietflyer There is no need to specify the country, it clearly says [faa-regulations] in the tags. See also Should questions include "tags" in their titles? on Meta.

@Bianfable -- tags can be (and often are) added by anyone, so there is no guarantee that they reflect the intentions of the original poster.  A question should stand alone on its own content, and should not depend on tags to clarify the intent of the question.  This is a different issue from whether or not tags should be included in the title.

@Bianfable -- I sense a ASE meta question brewing here.  But I can't support the idea that  a question should be dependent on tags for clarity, i.e. if you deleted the tags you wouldn't understand the question-- that is not adequate.

There is a subscription site called "Hein online" that allows searching of Federal Regulations back to around 1938 -- if there's a way to do this without paying, I'd like to know about it-- haven't found google to be very effective for older regs--

I've not been able to find copies of aviation regulations for the USA pre-dating 1938.
In the 1938 Code of Federal Regulations, Title 14 item 20.44 speaks of "specifying the weight and powerplant of aircraft which the applicant is deemed competent to pilot".  Classes 2M, 3M, and 4M are all "multi-engine" classes.  However, the word "rating" is not specifically used in relation to these classes.

Similar content may be found in the 1939 edition of the Code of Federal Regulations.    It appears that no update to Title 14 was published in 1940.
In the 1941 Code of Federal Regulations, Title 14 item 21.350 specifically references "aircraft class ratings".  Item 21.350(c) is "multi-engine land" and item 21.350(d) is "multi-engine sea".

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text:
How should this ODP be flown at KELP?
The KELP (El Paso) Obstacle Departure Procedure (ODP) says:

Rwy 4, 8R: climbing right turn heading 120° and ELP R-150 to 9000, then proceed on course.
Rwy 22, 26L: climbing left turn heading 120° and ELP R-150 to 9000, then proceed on course.

Do I fly from the VOR or to the VOR on ELP R-150?
From: would actually mean I use 150 and fly south east
To:   would actually mean I use 330 and fly north west
Do you have a reference for the answer?

In instrument flight, when a Radial is given, it is always the “From” indicated radial on your OBS. In vectoring, turning an aircraft at acute angles is avoided. Flying a Heading of 120° to intercept a Radial of 150°, your only option would be to turn toward a course of 150° degrees (Southeast) until reaching 9000 feet MSL. The field elevation is already almost at 4000 feet MSL. You only have to climb another 5000 feet before you can turn on course. The dead giveaway is the fact that a course of 330° would have you passing in or over the KBIF military Class D. Probably, before you could affect a frequency change

Hi.  Do you have an AIM ref or other for this?  "In instrument flight, when a Radial is given, it is always the “From” indicated radial on your OBS."
Thanks

@chup - In various locations in 14CFR like Part 71 & 73, radials are defined as from the NavAid. Per the Instrument Flying Handbook, a radial is defined as - “Radials. The courses oriented from a station.”

Furthermore, on page 9-10 of the IFH, it states, “The courses oriented FROM the station are called radials. The VOR information received by an aircraft is not influenced by aircraft attitude or heading. [Figure 9-10] Radials can be envisioned to be like the spokes of a wheel on which the aircraft is on one specific radial at any time. For example, aircraft A (heading 180°) is inbound on the 360° radial; after crossing the station, the aircraft is outbound on the 180° radial at A1. Aircraft B is shown crossing the 225° radial...”

“...Similarly, at any point around the station, an aircraft can be located somewhere on a specific VOR radial. Additionally, a VOR needle on an RMI always points to the course that takes you to the VOR station where conversely the ADF needle points to the station as a RB from the aircraft. In the example above, the ADF needle at position A would be pointed straight ahead, at A1 to the aircraft’s 180° position (tail) and at B to the aircraft’s right.”

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text:
When can an ASEL CFI with Commercial AMEL rating provide and log dual given?
Specifically
Does not hold MEI.
Has 0 hours in Cessna 310.  Can he provide and log dual given in that Cessna 310?
and,
What if he is not AMEL only ASEL?
Current regulation?
14 CFR 61.51(e)(3)
https://ecfr.federalregister.gov/current/title-14/chapter-I/subchapter-D/part-61/subpart-A#p-61.51(e)(3)

I'm not sure what your second paragraph means: "and, What if he is not AMEL only ASEL?" You already mentioned that the CFI is an ASEL CFI only, is there some other scenario you're asking about?

To me this question is similar to "Can a non CFII instructor provide instrument training"  which the answer should be yes, except he can't provided the training required for a rating.

As PilotDan has stated, instruction in the Cessna 310 can not be logged by either pilot unless the instructor holds an MEI. If either of the pilots does not hold a AMEL rating, that pilot can not log the C310 time as flight time at all. Not even as a safety pilot per 14 CFR 91.109.
Ad hoc training can be given by any AMEL rated pilot. It just can’t be logged as any type of flight time by anyone not possessing the proper category and class rating per 61.51. And, it can not be logged as instruction time without an MEI.

In either situation, the CFI can't log dual given and the other pilot can't log dual received, though they could be a safety pilot in the first situation.
The applicable reg is 61.195(b).

What if the training is not for a rating.   The non CFI is receiving training and it will not apply toward the rating.  All that training will have to be done by a qualified CFI.  Thanks

I think https://ecfr.federalregister.gov/current/title-14/chapter-I/subchapter-D/part-61/subpart-H#p-61.195(b) is the likely the correct answer.  It specifically states "Aircraft ratings" and I was thinking of it more like the "rating" used like "instrument rating"  which seems to have more grey area.


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Could airline tires be deflated in mid-flight to reduce weight?
As a means of weight reduction to reduce fuel consumption, could the tires on an aircraft’s landing gear be deflated once airborne then re-inflated just prior to landing?  Is that feasible and safe?

Comments are not for extended discussion; this conversation has been moved to chat.

No. Because they would still need to be inflated at take off, when the plane is already at its heaviest. So you'd make zero impact on the weight budget in the one instance where it would truly matter...assuming the intrigue here is considering how to squeeze a few more suitcases on there.

Easier to just chuck some landing (taxiing?) gear away on takeoff a-la the U-2's "pogos"

Uhm, thought I'd leave a little note for those who are confused as to why this should affect the weight at all (possibly including the OP too, in case he's also confused). It only matters because the tire has a higher air density than 1 atmosphere and because the outside air will be at less than 1 atmosphere. I neither of these were true, the inflation of the tires wouldn't change anything, since the buoyancy from the outside air would counteract the mass increase.

@Mehrdad Really? Buoyancy is based on the volume of the displaced fluid, a flat tyre's volume is not much less than an inflated one.

@Koyovis: "Buoyancy is based on the volume of the displaced fluid, a flat tyre's volume is not much less than an inflated one." Well, no, it's based on the weight of the displaced fluid. But in terms of your actual point, I don't see how that statement goes against anything I said. The upwards buoyant force is obviously the same as you say, and yet the mass of an inflated tire is larger, hence it weighs more.

@Mehrdad yes it does indeed, no argument there.

It's a good question. Estimations of 120 kg weight savings of compressed air in the tyres have been made, that is not inconsiderable. A detailed estimation with dimensions from this link. The B777-300 has tyres size 52 X 21 R 22:

52 is outside diameter in inches = 1.32m
21 is tyre wide in inches = 0.53m
22 is rim diameter in inches = 0.56m
Side area of the tyre itself = $\frac {\pi}{4} (1.32^2 - 0.56^2) $ = 1.23 m$^2$.
Volume of the tyre = side area * width = 1.23 * 0.53 = 0.65 m$^3$. The B777 has 12 of these wheels so total tyre volume = 7.8 m$^3$.
Air at 1 bar pressure = 1.225 kg/m$^3$. Twelve wheels at 16 bar contain 7.8 * 1.225 * 16 = 153 kg of air.

Weight savings in the order of magnitude of 150 kg, every trip for two to three decades, are nothing to sneeze at. Consider it the other way around: how sane would a person be if they would propose to put an extra 150 kg permanently on board because the extra fuel would be inconsequential?
To re-inflate the tyres, all infrastructure is on board already: jet engine final compressor stage pressure is definitely higher than the required 16 bar. From Rollys Royce The Jet Engine issue 5:

After cooling off from 700C - 1000K, to 30C = 300K, that is in the order of magnitude of the required air pressure for re-inflating the tyres. So technically, all is possible. The tyres can be re-inflated with existing infrastructure on board. Of course not with the existing unmodified infrastructure as it is now, it will need to be tweaked, but order of magnitude wise everything is on board already. Tweaking a tyre pressure inflator in the 21st century is a much simpler task than inventing the first autopilot in 1930. I mean come on guys.
Aluminium 1inch tubing weighs about a kilo a meter, and the engines are very close the wheels. 10 m extra tubing would be a lot. The hydraulic oil has a heat exchanger with the fuel in the wing, which is super cold just before landing, having spent all this time at -30C. Let's say that the total re-inflation installation weighs 30 kg tops.
Where the whole idea falls apart is in the safety aspects. We're approaching an airport with low fuel reserves and with flat tyres, too many things can go wrong and Murphy pops his head up from every corner. The undercarriage is a considerable weight and contributes virtually nothing to the flying and economic capabilities of the aircraft, but we better make sure that it pops out every time and can cushion the last stage of flight before all passengers can get safely off board.
In air passenger transport, if safety needs to compete with economics, safety wins every time.
@mikerodent suggests in a deleted post that the tyres could be filled with helium. Not a bad suggestion actually: helium weighs 0.164 kg/m3, weight saving for the B777-300 is 132 kg and the tyre does not need to be re-inflated. The answer deserves to be undeleted and accepted.

to those 150kg you have to subtract the weight of the extra redundant plumbing and heat exchangers before you can claim "Weight savings in the order of magnitude of 150 kg". You have done all the calculations for one side of the equation, you can do the same for the other. Alternatively, you can check if the fuel saving of those 150 kgs compare to the extra engine losses incurred in bleeding HP air.

@Federico yes more details need to be worked out before it can be implemented, and total weight savings might be less. But that is beside the point: it won't be implemented no matter how attractive, because it imparts on safety.

https://www.grainger.com/category/aluminum-tubing/tubing/pipe-tubing-and-fittings/plumbing/ecatalog/N-qwo#nav=%2Fcategory%2Faluminum-tubing%2Ftubing%2Fpipe-tubing-and-fittings%2Fplumbing%2Fecatalog%2FN-qwoZ1z0lqhpZ1yzahygZ1z0o05j resist only up to 400°F, less than the 500°F (and more) found in the HP section of the engine [https://aviation.stackexchange.com/a/11483/1467 <- note that here we don't even get enough compression] I'm not sure that Aluminium would be acceptable, nor that a non-redundant system would be ok. So 30kg would be the lower limit, not the upper.

@Federico sure man. The stainless steel version of tuning used for hydraulic systems does not weigh much more.

"In air passenger transport, if safety needs to compete with economics, safety wins every time" I believe it's not true, counterexamples easy to list: Aircraft disappear because they aren't tracked over oceans, and then huge amounts of public money are spent to SaR them. Aircraft use achaic data recorders with archaic ULB to (not) retrieve them. Aircraft can legally fly without a system to alert the crew when they are landing with gear up, etc. Airlines shareholders have a strong lobbying power on regulations.

@mins, sorry man, cannot follow that. Yes there is old stuff on board aircraft, even 80186 processors, and that is because old = known, new has safety risks. None of the old stuff you mention has been implemented because it is cheaper than better new stuff.

"None of the old stuff you mention has been implemented because it is cheaper than better new stuff": I'm not challenging past actions, but emphasizing that status quo attitude is cheaper than actively improving security. The examples I listed are status quo cases made possible by the influence of airlines facing competition cost challenges.

You want to reduce weight by removing a few kilos of air, and adding an equal, if not greater, amount of kilos of pumps needed to reflate the tires (see David Richerby's answer for details), not to speak the additional costs incurred for periodically checking the correct functionality of said pumps, that would be marked a "safety critical" (because a landing with deflated tires would have really bad consequences) and require being checked quite often.
So, feasible maybe. Safe not really. Meaningful I don't see how, particularly when there is the risk of tire burst, as shown in this video linked by March Ho.
A lighter alternative would be to use bleed air from the high pressure section of the engine, but it would still probably(*) be heavier than the air you are pushing out (you need heavy duty pipes to transport high pressure air from the engines to the wheels), and not a good solution safety-wise: what would happen if you have to land without engine power?
(*): as user3528438 mentions in their comment, it would depend on the wheel size.
Additionally, this Goodyear manual provides insight in some other problems: (emphasis mine)

ALLOW 12-HOUR STRETCH AFTER MOUNTING

All tires, particularly bias tires, will stretch (or grow) after initial mounting.
This increased volume of the tire results in a pressure drop.
Consequently , tires should not be placed in service until they have been inflated a minimum of 12 hours, pressure rechecked, and tires re-inflated if necessary.

It means that the tire needs adjusting after being inflated, possibly removing the possibility of inflating and deflating it continuously (or, alternatively, you have to reinflate it at least 12 hours before landing). Note that this manual is written for current operations, that call for a replacement/inspection of the tire if pressure drops below 80% of nominal value. Procedures should be rewritten for inflation/deflation cycles, and this could be a factor.

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Some military transport aircraft can indeed adjust tire pressure from the cockpit for soft field operations. The Antonov An-22 would be one example. A central tire inflation system is a standard feature on Soviet military trucks as well, and also used on some US trucks.
That the feature was removed on later versions of the An-22 should tell you something already.
Yes, a mass saving of a few dozen kg on a large transport aircraft is feasible, but you need to add pumps and plumbing. In order to clear civil certification standards, this system must be redundant, so you need to have most components duplicated. Also, the energy stored in the pressurized tire cannot easily be recuperated, but needs to be subtracted from the calculation of possible fuel savings because it must be mustered for re-inflation.
Without a more detailed investigation I would expect that the mass savings are minimal and do not outweigh the increased complexity and added failure modes.

I bet An-22 had those to adjust the tires for the runway at hand (paved - high pressure vs. unpaved - low pressure) rather than save on weight.

@Agent_L: Yes of course. That it was not done for weight saving is obvious from the fact that none of the Soviet airliners used this system.

Along with the pump (and such) involved, it should be noted that inflating the tires of a large aircraft is a fairly non-trivial undertaking from a safety point of view.
In particular, when you're inflating a tire you normally put it into a tire cage, like this:

Photo Credit: Martins Industries.
That's not the only variety, but you get the idea. It's nearly certain that the cage you'd need around the tire to inflate it safely would, all by itself, be considerably heavier than the air inside the tire.
Just for what it's worth, there are only a few tire cages that can be used without removing the wheel and tire from the aircraft--and if I'm not mistaken, they tend to be even larger and heavier than the more common ones like I've linked above (though in fairness, not drastically larger/heavier).

I saw the cages and looked for videos of airplane tires exploding.This video makes it pretty clear why cages are necessary

And the fact that one needs these cages implies that there's a decent chance that trying to inflate the tyres in flight will lead to you trying to land with fewer tyres than you took off with.

@DavidRicherby I'd be more worried about less fuselage.

You simply wouldn't save any weight.
One of Goodyear's largest commercial aviation tyres is the 54x21.0–23 Flight Leader,* used on Airbus A330s and A340s. That is, outside diameter 54 in, width 21 in, rim diameter 23 in (137x53x48 cm). Assuming the tyre has rectangular cross-section, each tyre has a gross volume of about 0.7m3. This is an over-estimate of the air volume, since the cross-section is curved and the calculated volume includes the rubber.
These tyres are inflated to 235 psi, which is about 16 bar. The density of air at 1 bar is about 1.2 kg/m3, so the total mass of air in a tyre is about 16 kg.  An A330 or A340 has ten tyres (I guess the nose wheel tyres are smaller but let's neglect that), so the total mass of air is at most 160 kg.
The maximum take-off weight of an A330/A340 is in the ballpark of 250,000 kg, so a 160kg weight saving is something in the region of 0.06%.  Another way to see the scale of it is that 160 kg is basically two passengers with no checked luggage, or about 500 g per passenger (these planes carry about 300 people).
However, you would also need to install a multiply redundant high-pressure air compression and distribution system.  A quick Google suggests that the sort of air compressor one might use in a workshop weighs 30–40 kg, and only does 8 bar (remember, we need 16). Probably most of that weight is the reservoir tank and I'm not sure how this compares to something that would be needed to inflate aircraft tyres. However, you're going to need something pretty beefy to pump seven cubic meters of air at sixteen bar in reasonable time, and even four of those ordinary workshop pumps would eradicate most or all of our possible weight savings.

* Other brands of aircraft tyre are available; I have no links with any manufacturer.

I'd wager that even the air hoses and fittings capable of 300 psi and moving a large amount of air (3/4" ID at least) would weigh multiples of the weight of the air "saved" by deflating the tires.  Also, air compressors get very hot and use a lot of energy.  Maybe some energy could be reclaimed from deflating the tires tho...

"You simply wouldn't save any weight." And then you compute that 160 kg is saved. There is compressed air on board already,from the engines. Those compressors you find have not been designed for airborne use and are produced with no regard to weight.

@Koyovis What about engine-out landings? And will the engines provide enough pressure? The RR Trent 700 (as used on the A330) has an overall pressure ratio of 36:1, which I guess means you can bleed off air at 36 bar, when you're close to sea level and at full power. But you're going to need to reinflate the tyres as you're coming in to land, so presumably at greatly reduced power. Is there going to be enough pressure available?

Yes the safety issues and emergencies are the problem. Re-inflating via the bleed air isn't, simply re-inflate before final descent.

@Koyovis Even using engine bleed air, you would have to install extra plumbing from engines to the wheels, that would have to be redundant.   That would add weight.  In addition, the air would have to go through a fitting that allowed the wheel to rotate 360 without losing air.  If a tire exploded in the air during inflation you would probably lose the entire airframe and everybody in it.  Pieces of the tire could easily damage the wing sufficiently for it to fail in flight, leading to loss of the entire airframe.

@Flynn The rotating inflation gadget already exists. And, while I agree with your objections, they're covered by the other answers.

""You simply wouldn't save any weight." And then you compute that 160 kg is saved".  Correct, within the margin of measurement error you are saving no weight.  Engineering is not platonic arithmetic.

@David Richerby Yes, I knew they existed.  My point was only that it was just more weight and complexity and something else to fail.   The whole idea of inflation in flight is so silly one hardly knows where to start.   But the problems have been well discussed including by you.  And one learns a lot from the answers and comments even if the original proposal was far-fetched, so in many ways it was a good question.

Everyone here is missing one critical fact - airline tires are filled with Nitrogen. The friction of the tire on tarmac could raise the temperature enough for the Oxygen to react with the rubber, causing a fire. Nitrogen is mostly inert, making it much safer.
The entire question is irrelevant because you would be unable to inflate the tires again with only Nitrogen unless you carried that same volume aboard the aircraft in a second tank, reversing the weight savings.
Alternately, you could carry the lab equipment needed for fractional distillation, but I assume that would be even heavier.

You know that the fuel tanks are also filled with nitrogen enriched air, and that is indeed created onboard, don't you?

@PeterKämpf as already mentioned in other comments on my answer, that system is not really comparable. 88% nitrogen is a bit far from the 99.X% in the tires

@Federico as it is now, yes. Obviously, the tyre inflation system would be re-designed to provide a higher percentage of nitrogen.

@Koyovis and have you checked how much heavier and bulkier this redesign would be?

@Federico those are design parameters. The system could come out heavier and bulkier, or the present system could do it if given more time, or a lightweight oxidiser solution could be put in place. Initially before the design, everything is too heavy because it is not designed yet for low weight.

I read recently about a proposal to replace the gas in aeroplane tyres with helium to save weight.  This would not require re-inflation in flight but would require better gas barrier layer inside tyres to prevent faster gas diffusion.

As Cody already said, "Alternately, you could carry the lab equipment needed for fractional distillation, but I assume that would be even heavier."  Couldn't be more straightforward.

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Can lasers be used for weather detection along an aircraft's flight path?
Aircraft use RADAR for weather detection such as storm clouds, rain, etc. Why don't they use LASER instead to measure the density of the weather ahead?
RADAR has been known to return false signals of weather ahead (i.e. clear cloud, low density).
Would LASER be a more accurate application as LASER has been used in many other areas such as measuring terrain topography, density of some materials, temperature, etc.
P.S  Please forgive my ignorance if this technology is already in use for this application.

LIDAR are generally on the ground, but as this answer says, there are experimentations for airborne LIDAR to detect CAT.

not at 35,000ft looking for weather hazards...LASER operates in many wavelengths of light.Some of which is not visible to the eye without equipment and media such as glasses and or vapour such as smoke,etc.

Related

I'm voting to close this question as off-topic because this does not appear to be relevant to aviation weather.  Such a question is general meteorological science.  To make sure I am completely clear, this would not be unlike asking a materials science question simply because it may be incorporated into the body of an aircraft.

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text:
How does the less dense air affect the reciprocating aircraft engine performance?
How does the less dense air can affect the reciprocating engine performance ? For an example " rotax engine "

Lower air density means less compression, hence less mean effective cylinder pressure during expansion, and thus less power, since shaft power is proportional to the product of mean effective pressure and angular speed...

Thanks I appreciate your help


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What plane is this instrument panel from?
I came across this instrument panel being sold as a potential craft project. I instantly knew what I wanted to do with it. I am going to give it, filled with old pics to my father-in-law who used to be a pilot. It would, however, make for a better story if I can find out the type of aircraft this instrument panel came from. Can anyone identify it for me, please? Maybe even just a hint at where to look?

You've come to the right place; if this is from a mass-produced aircraft some wise enthusiast here will probably be able to tell you. But be aware, sometimes instrument panels are just made bespoke for a specific aircraft and its almost impossible to tell which one it came from. If there are any markings and/or serial numbers on it, that might help identify it, also the dimensions may help narrow it down.

Things that stand out to me:  the "Inverter Power" and "Radio Master" labels on the right, and the various audio switches along the top being part of this main panel instead of on a subpanel.  I think the vertical notch right of center along the bottom is for a gear lever?  Possibly a twin, owing to all the gage holes and the doubled-up holes on the left, but it seems pretty narrow.

The existence of DME/NAV switches indicates to me this is a post-war aircraft? Can anyone confirm that is right?

The existence of DME/NAV switches indicates to me this is could be a WW II aircraft that has been updated  ;)

In what part of the world was it found? Can you provide measurements? Could even be a four engine aircraft with all those instrument holes....

Is there any part number on it?

It was in use, there are traces left from components that have been there.

Potential lead: the pointed top is typical of some DeHavilland planes. The Beaver specifically has one, but this panel doesn't seem to match that.

Try and post it here.... https://www.pprune.org/aviation-history-nostalgia/547101-what-cockpit.html

It is definitely similar in shape to the Dehavilland Dove or Heron, but there are still plenty of differences in the instrument cut-outs.

Interesting that the pilot/copilot stacks are significantly assymetric.  They must have had significantly different cross checks.

@tmptplayer or it's from a single operator aircraft. Light aircraft tend to have only one primary instrument stack, the other cutouts could be for navaids, backup engine instruments, etc. etc.

Single pilot was my thought or single placement in a helecopter.

From your second picture I can read:
XXXX Deice Deactivated
Oxygen System Deactivated
TKS Wing Tail XXXX
Deactivated
Fuel Transfer Deactivated
Autopilot Desctivated
Heater Inop
Although from looking at the other pictures one would think there is much more that is deactivated on this aircraft.

Makes me think it was in a bone yard. Pos mikitary?

just removed one guage and it says "Prop Ice Guard Amps". Also on the back i found to mounts that say "P/N 150 PL-12 AN800D10" but all that told me is it is a standard antivibration mount from LORD Corp

And probably the last bit of info is what appears to be an inspection stamp on a bracket on the back. see image

OK. de Havilland DHC6-100 1967 has the closest instrument panel i can find so far. it even has the strange risers on oll the screws. BUT NOT EXACT YET.

That's an incredibly neat craft idea I now want to steal.

I was so close. I was looking at the de Havilland Otter but a mid 50s. A friend finally found it. It is a mid 70s DH-114 de Havilland Riley. Thank you verandaguy for leading to the de Havilland. That was most helpful. Now I can finish the Xmas present and be able to tell my Father-in-law the exact type of plane it came from. Cheers.

A picture of the cockpit would be really helpful.


text:
How do I choose the correct params?
I am very new to airplane design. I am working on this project to design a flying wing glider (no propeller) that would glide in a circular pattern with a fixed radius of 250 m after being released at a height.
Here is all that I know now:
I was reading around the internet that a choice of delta wing would be good for maneuverability of the glider, however it would be bad for operating at low speeds. I guess that this design would not fit me because it has very poor lift. But at the same time a wide plank kind of design would not be able to perform a turn that well. So I've decided to choose a swept wing for the flying wing. However, I am not able to decide properly the angle of sweep required. I know that the stall angle and the twist of the wings depend upon it, but I don't really know how to tackle the problem (as I guess I would require a high stall angle to provide me the lift).
When it comes to choice of airfoils, I am currently looking at HS522, MH60, and NACA 22112 and Eppler 186 airfoil. However, again I can't really choose between them. From my understanding, I require a reflexed airfoil with high camber and high thickness to provide me with good lift as I plan to operate at slow speeds close to ~20 m/s.
Wing loading and banking of the turn is another thing that I understand is related to each other but I don't understand how I can reach a common point between them. From my preliminary analysis of the situation, I have reached a bank angle of 9°. Is that a good angle? I read on online resources that the standard bank angle is 25°. Is that relevant in this case? Or do I approach this problem by deciding the wing loading and then move on to the bank angle?
Thanks

Why does it have to be a flying wing? What ist the purpose of circling? Thermals? Should it simply not fly away?

@PeterKämpf Well the idea was to create a glider that could perform a circular (helical) turn around a park if it was released from a high tree. Was planning to improve it to work for a forest where we have massive trees. I don't think so that it will be able to get any thermals. It just has to stay in air for sufficient time.

A flying wing needs an airfoil with an s-bend, so no regular airfoil (none of the ones you mentioned). And you need to incorporate a significant twist towards the wing tips to make it fly stable (incidence is about 10° lower at the tip).
Knowing your turn radius and your desired speed your bank angle is a given, independent of aircraft mass. You can compute it with the centrifugal force and the lift portion that points into the turn...
r = TAS * TAS / ( g * sin( bank ) ), with true airspeed (TAS) in m/s, g = 9.81 m/s/s, r in meters. (if I'm not mistaken)
When you know the mass of the aircraft (roughly) and your airfoil lift coefficient that you want to work with (e.g. 0.5 to 0.8 or so) you can work out how much wing area you'll need to fly that turn without stalling. You could also chose a wing loading and compute the lift coefficient for your turn using the basic lift equations and the force equilibrium...
25 degrees is a good bank angle for passenger aircraft, more makes the passengers scared :) Not applicable here.
I'd start with an existing design of a flying wing and then make alterations to that. A famous aircraft that comes to mind is the Horten H.IV glider. Another example could be the Marske Pioneer 3

Any recommendations for the choice of airfoil? I saw that the Marske Pioneer 3 uses M-35A. Thanks


text:
What Is this on the fuselage of The Fokker F100?

What Is  this on the fuselage of  The Fokker F100?

Does this answer your question? Where is the inlet for the APU on a Fokker 70/100?

Yes it's the apu exhaust ... there is a post about it:
Where is the inlet for the APU on a Fokker 70/100?


text:
What is that A340 cockpit panel on the glareshield?
I want to know the panel's name and how it works. This picture is an Iberia Airbus A340-600

That display is part of the AFS (Auto Flight System) Control Panel on the FCU (Flight Control Unit):

(Airbus A340 FCOM - Auto Flight - General)
The general idea of the FCU is that you can push in the knobs, which will put the respective autoflight mode in an automatic setting (called managed mode) and show dashes in the display, and you can pull the knobs out and then rotate them to manually overwrite values (called selected mode).
In your example, the display currently shows:

SPD 250: The autothrust system is maintaining a manually selected speed of 250 knots.
HDG ---⚫: The lateral autopilot mode is in managed mode, meaning the aircraft will follow the flight plan programmed into the FMGS. See also this answer.
ALT 08000: The current target altitude for the autopilot is 8000 feet.
V/S -1000: The vertical autopilot mode is to maintain a manually selected descent of 1000 feet per minute.

The FCU allows you to switch between selecting a heading (HDG) vs. a track (TRK) and a vertical speed (V/S) vs. a flight path angle (FPA) with the button at the center. The HDG and V/S in the center of the display mean that the current mode is in heading and vertical speed.

thank you very much


text:
Reference Map of EUROCONTROL charging zones
I'm building out an application to calculate EUROCONTROL navigation fees. EUROCONTROL make one of these already, but I need to replicate the functionality within my own stack.
In the aforementioned EUROCONTROL app, there is a data file containing what appears data in a geospatial format. In order to do the required calculations, I first need to understand that data format.
My first impression is the first row is some metadata. EG means the UK. From there the number pairs begin and end the same, indicating a closed polygon.
Example for EG (so UK) is here
3274 -600
3300 -600
3360 -600
3420 -600
3480 -600
3540 -600
3600 -600
3660 -600
3660 -540
3660 -484
3660 -480
3660 -420
3660 -390
3660 -360
3660 -300
3660 -275
3660 -240
3660 -180
3660 -120
3660 -111
3660 -60
3655 0
3600 0
3420 300
3300 300
3090 120
3067 120
3060 88
3040 88
3000 -15
3000 -120
2930 -480
2940 -480
2970 -480
3000 -480
3060 -480
3140 -330
3235 -330
3265 -490
3320 -415
3325 -440
3320 -495
3285 -540
3274 -600

Plotting the numbers in an SVG, so simply turning the numbers into pixels gets me this...

I'm hoping that matches a EUROCONTROL map someplace. Geospatial data  can be quite complex, and I don't really understand a lot of that complexity, so projection might be off (or any number of other things).
Can anyone point me in the direction of this within an official EUROCONTROL document? From there its a short hop to understanding how that original data file can be translated into some GIS software that gets me to the next stage of my project.

That is the map of the European airspace, however you have mirrored it such that north is now south and vice versa.

source: Eurocontrol

GC    Spain Canarias
AZ    Portugal (Azores)
LP    Portugal Lisbo
LE    Spain Continent.
LF    France
LS    Switzerland
LI    Italy
LM    Malta
LG    Greece
LC    Cyprus
LT    Turkey
UD    Armenia
UG    Georgia
LB    Bulgaria
LA    Albania
LW    Republic of North Macedonia
LY    Serb.-Montenegro
LD    Croatia
LQ    Bosnia Herzeg.
LJ    Slovenia
LO    Austria
LH    Hungary
LR    Romania
LU    Moldova
EP    Poland
EG    United Kingdom
ED    Germany
EH    Netherlands
EB    Belg.-Luxembourg
ES    Sweden
EK    Denmark
EF    Finland
LZ    Slovak Republic
LK    Czech Republic
EY    Lithuania
EN    Norway
EI    Ireland
EV    Latvia


text:
How are aviation sectionals and enroute charts created?
Who exactly handles this?  Is this done by the US Government via one of its printing offices?  Or is this private?  I know Jeppesen does some charts, especially approach plates, but I am not totally sure how all this is done.

Are you asking who creates and publishes the charts, or how they are actually made? In case of the latter, this is a duplicate: https://aviation.stackexchange.com/questions/38683/how-are-aeronautical-charts-made

Both who manufactures them and the process by which it's done

“Or is this private?” - with the current political state, we’re not very far from it.

I believe the FAA got out of the printing business by implementing a new business model for charts where they license the product to private industry who then take on the risk/reward of the printing process for their customers.
From the FAA AIS website:

The Federal Aviation Administration (FAA), Aeronautical Information Services (AJV-5), is transitioning aeronautical charts and related products print and distribution operations from existing contract(s) to multiple no-cost agreements with private industry. Digital product files allow industry the opportunity to print, sell, and distribute aeronautical charting products to the public in an Available on Demand (AOD) approach. Any party interested in the opportunity to become an approved print provider should submit product qualification samples from the files below (in accordance with instructions in the Print Provider agreement).

More detailed information about the program is also available through the FAA contract award management site via an official announcement. So to answer your question, the printing of charts is now handled by private industry, not the FAA.
Since you mentioned Jeppesen specifically, note that they are a bit of a different concept because they take raw data from each state's official AIP and then integrate it together in their own way and perform other value-added services. They then print their own end-product for their customers.  It is based on the same raw data but their product should not be considered the same as what is produced by the FAA.

The art and science of creating maps is called Cartography. The experts in the field, Cartographers, are educated for the same  with degrees in Geography. There are many schools that offer such programs.
The FAA, which falls under the DOT (Depart of Transportation), carries out a tonne of duties. "Air Traffic" is one of them, and creating & publishing charts are duties of the offices that handle the Air Traffic section.
For a complete hierarchical breakdown of the FAA (just the FAA, not the entire DOT), refer to the Understanding The FAA page on DOT's website which also contains an organization chart.
That said, one of the offices[a] under the Air Traffic Sector under the FAA has a department that employs Cartographers and other agents to create and work on these maps. How big is the department, and how do they "actually" create these maps (what software and machines they use, how are they supervised etc.) are questions that can't be answered.
[a] FAA ▸ Air Traffic ▸ Aeronautical Information Services Group ▸ NFDC ▸ Charting Groups/Teams.

As far is printing and publication is concerned, the previous answer covers it.


text:
Is lift in fact a kind of drag?
Given the similarities between Lift and drag and their very similar relationships to the density, surface area, and airflow velocity, could one, in theory, describe aerodynamic lift as a kind of drag where most of the force from lift is just directed in the vertical direction where this vertical component of the force opposes gravity?

It is, in the same way that drag is effectively lift in the direction you (often) don't want it to be.

Given the similarities between music and noise and their very similar relationships to the ear, pressure waves, and psychoacoustics, could one, in theory, describe music as a kind of noise?

Shoegazer here. We certainly think so!  @PhilFrost

Related: https://aviation.stackexchange.com/questions/27885/is-induced-drag-essentially-nothing-more-than-a-specific-type-of-form-drag

An 'aerodynamic force' (just one force...) appears when a body is immersed in a fluid stream. By convention, two components are chosen, one of them parallel to the stream direction, called 'drag', and the other one, perpendicular to that 'drag' is termed 'lift'.

such a simple (and correct) answer

Yes. Just to be more accurate: lift and drag are just components of the force vector but they doesn't say anything about the nature of the force (inertial or viscous for ie.)

At least as the terms are normally used, lift is not drag, and drag is not lift.
What I suspect you're looking for is: "induced drag", which is the part of the drag that's caused by an airfoil generating lift.
Induced drag is opposed to "parasitic drag", which includes things like friction drag (friction between the airfoil's skin and the air) and interference drag (e.g., there's some extra drag where the wing mounts to the fuselage) and so on.

Using the wing as a frame of reference, then the aerodynamic force is related to diversion (change in direction) of the relative flow. In a normal situation, the relative flow in front of the wing is horizontal, and the wing curves the flow downwards a bit. Assuming an ideal case where the flow speed isn't changed, then the horizontal component of the initial velocity is decreased somewhat (induced drag), and the vertical component is increased from zero to some non-zero value (lift). In a real world situation, the flow speed is reduced due to friction, viscosity, profile drag, ..., which further reduces the horizontal component of initial velocity (other components of drag).
Using the air as a frame of reference, drag is related to the forward acceleration of air, while lift is related to the vertical acceleration of air. Consider the case of a bus on a highway, the air is accelerated forwards, most of this due to the low pressure area behind the bus, some due to the air being pushed forward and around the front of the bus, while there is zero lift.
On a side note, lift calculations are simpler than drag calculations because lift calculations can be based on flow field calculations that calculate speeds (relative to the wing) and the coexistent pressures related to those speeds. Drag calculations are more complicated because they need to take turbulent and boundary layer effects into account.

One might ask what application would considering lift as a type of drag have?  They are compared as L/D ratio as a parameter of design performance, however,
especially as pilots, knowing the distinction may
be extremely important as they apply to flight.
The amount of lift one can generate is dependent on
velocity (for a given aircraft).  Why this is critically important is that velocity is the kinetic
energy state, which is created by adding energy to
the system from the thrust.  A little
like filling a barrel.  One spigot is altitude
(potential energy) that you can pour back in as needed.  The other is drag.  Thrust keeps filling
the barrel, but it must be filled to a certain minimum
level or the plane does not have enough velocity
to fly.
Empty the barrel too fast (steep climb), you can grab your
potential energy (altitude) and pour it back in to
get flying speed back.  The thrust can only be added
at a certain rate, as we know on our take off roll.
Drag finds application as how much energy is being
taken out (as related to fuel consumption or max velocity) and can be compared with lift for a given configuration, AoA, airfoil type, etc.in engineering theory, but as far as piloting, not related IMHO.
Bob

Maybe the best answer here is that, while lift and drag are separate quantities, their behavior is governed by the same physics ie conservation of momentum and Newton’s Third Law of Motion.

Physics laws are universal, but lift and drag are 'separate quantities' only by convention. The aerodynamic force is decomposed in lift and drag just because it's useful to do it so. Other de-compositions are possible, if probably not as useful, but that is the only 'reality' of lift and drag, an instrumental reality...

I think that's the wrong way to look at it.  The lift force is 90 degrees to the airstream.  When you're gliding, the lift force is producing thrust, not drag, or else you would stop dead and fall from the sky.
I think of the lift-related drag component as the energy consumed in inducing the air to move from point A to point B, creating the pressure differentials and down thrust from action/reaction, as the wing passes through it.

No a component of the gravitational force acts as thrust in a glide, not lift.

But the gravitational force is straight down, not forward.

In a glide, a component of the gravitational force now acts parallel to and opposite the direction of, the relative wind, causing the airplane to move forward.

So when a helicopter tilts its rotor and angles its lift vector forward, the forward thrust component is generated by gravity?

No.  That’s not the same thing.  You need to do a free body diagram of that taking into account the relative wind as well.  Both  Lift and thrust there are being generated by engine power.  Now if the helicopter loses engine power and begins an autorotative descent, then the forward thrust is being provided by gravity.

A forces & velocities diagram of a gliding condition here https://imgur.com/DUJqHqo

They're actually just opposite sides of the same coin.  This is like arguing how many angels fit on the head of a pin.

No it is not. By definition lift cannot point forward as thrust (lift must be at 90deg to flight path). Claiming otherwise only disseminates confusion in this world of my-ignorance-is-just-as-good-as-your-phd

Sooo.. if lift is 90 deg to the flight path, and the flight path gradient is down, say 10 degrees, as in a glide, the lift vector isn't tilted forward 10 degrees and doesn't this tilt represent a forward component?

i can see why this is disorienting: if the flight path is down 10 deg, then yes: the lift is also tilted 10deg. It is not  pointing “forward” (even though it is now skewed 10 deg to the left) because your “forward” is now not left, but downwards-left... therefore NONE of the lift is contributing to moving you “downwards-left”. Mind you, the gravity has NOT been tilted 10 deg, so now a part of it IS actually pulling you “forwards” ... as it is pulling you “down”( which is a part of “downwards-left” that is your flight path) ... ok, maybe this is not clearer: just draw a picture

better yet, use the picture from Pilohead’s (also wrong) answer. See how none of the L is pulling you towards V?(they are perpendicular) But some of the W is pulling towards D? That picture is for a climb, where weight adds on to the Drag to pull you backwards. In a descent, part of the weight (W sin y) will add forward with thrust (if any) to pull you “forward” from the plane’s point of view

Ok I get it.  Thanks.

Why don't you ask a question?


text:
What are the correct procedures for military aircraft interception over international waters?
I recently read this list of errors in the film Top Gun.  The list is long and fairly well done, but there is one I realized when thinking about that movie and that is Maverick may have actually caused an international incident at the start of the film that precipitated into the climactic dogfight at the end.  That is, they intercepted the MiGs at the beginning of the film in an unprofessional and hostile manner which is not used when intercepting another aircraft over international waters.
Aircraft interceptions are a common occurrence in international airspace but there are specific procedures involved and protocol used to avoid an international incident.  Typically a two ship flight will maneuver alongside an unknown aircraft to visually identify it.  Any additional actions ie creating a collision hazard from a head on cross or hard maneuvering onto an aircraft’s six then tracking it with a fire control radar, as Maverick does at the start of the movie can be interpreted as a hostile act and could actually warrant use of force.  Therefore Maverick may be the aggressor here and the second MiG maneuvering for a shot at Cougar and lighting up his F-14 with its fire control radar would be a reasonable response to the F-14 two ship’s aggressive act over international waters.
Goddamnit, Maverick!
Soviet aircraft routinely approached US carrier battle groups during the Cold War, in some cases coming within 1/2 mile from the battle group.  They were intercepted on a regular basis but never in such an aggressive manner as is shown in Top Gun.  Unless said Soviet aircraft were acting in a provocative manner with posed a direct threat to the ship or the air wing aircraft, there really wasn’t much you could do as you were in international waters and subject to international law.  Doing otherwise would have been an act of war.
So am I wrong in this assessment?  Is the intercept of an unknown aircraft as shown in the movie uncalled for and hostile to the MiG pilots who were only engaged in their own maneuvers over international waters?  Can anybody shed further light as to how the USN would professionally intercept a bogey over international waters?

It might not be hostile, but it is bloody dangerous closing at each other head on at same altitude with 500+ kts each. If you spot the other plane at 1 km away, 2 seconds later you might already have hit each other.

The F-14 and its AIM-54 Phoenix missile was originally built for a particularly unfriendly intercept of Russian bombers from 100nm away in war time.
In peacetime, the TU-95 Bear bombers still needed to be intercepted.  That practice, at least in the north Atlantic, was usually for one of a pair of F-14s to fly directly toward the intercept point and well above it, while the other was several miles in trail and on the side that would be to the rear of the bomber(s) after intercept.  The first would overfly and make a descending turn to end up about a mile in trail, holding position there while the second did the same thing.  This ensured that (depending on ROE) weapons could always be directed at the bomber by one or the other F-14's throughout the approach.
Sometimes the F-14s were tasked with (really, permitted to) close with the Bear to "gather intelligence".  This resulted in lots of great pictures. You'll note that usually the F-14s in those didn't carry Sparrow or Phoenix missiles so they'd constitute less of a threat while closing (there were exceptions, though).  The Bear is about the size of a B-52, and depending on the day's plan you could come within a couple (Bear) wingspans of one. This in theory was risky (they had a rear gunner who's arc of fire you were to avoid) and took it out of your missile envelope, but you did have a gun...

As the Bear got closer to the carrier, more F-14s would arrive and things would get a little spread out for deconfliction.

But everything was smooth, slow, and initially done from a significant distance. Nothing at all like the Top Gun intercept, which would have permanently grounded all four crew members.

The manoever in the movie is risky and unrealistic. What do you expect from Tom Cruise?  ;-)
There is indeed a Manual concerning Interception of Civil Aircraft governed by ICAO. It says:

4.1.2.1 The following method is recommended for the manoeuvring of intercepting aircraft for the purpose of visually identifying a civil
  aircraft:
Phase I
The intercepting aircraft should approach the intercepted aircraft
  from astern. The element leader, or the single intercepting aircraft,
  should normally take up a position on the left (port) side, slightly
  above and ahead of the intercepted aircraft, within the field of view
  of the pilot of the intercepted aircraft, and initially not closer to
  the aircraft than 300 metres. Any other participating aircraft should
  stay well clear of the intercepted aircraft, preferably above and
  behind. After speed and position have been established, the aircraft
  should, if necessary, proceed with Phase II of the procedure.

There are briefing documents for pilots who might find themselves being intercepted by Air Force or law enforcement for any reason. For example:

In-Flight Intercept Procedures  (FAA)
Intercept Procedures (CFI Notebook)

The basic idea is that both, interceptor and intercepted aircraft always act in a predictable manner.
Now, how much the Air Force or law enforcement will follow or deviate from this standard approach might depend on the specific situation and - just my guess - personalities.

Somewhat useful as background for a peaceful intercept, but a manual for intercepting civil aircraft is not really relevant to the question.  The question concerns the rules of engagement for Navy aircraft responding to a perceived military threat against the carrier group.

The question is about common occurrences in international airspace, and about powers that are not in war with each other. I am looking forward to more suitable answers.


text:
Airspeed change in wind shear
The instrument questions I’m studying say that a decreasing tailwind will increase airspeed.  I can’t wrap my head around why this would be the case.

Let's consider a simple example: you fly at a true airspeed (TAS) of 100 kt with a tailwind (TW) of 20 kt. This would result in a ground speed (GS) of
$$ \text{GS} = \text{TAS} + \text{TW} = 100 \, \text{kt} + 20 \, \text{kt} = 120 \, \text{kt} $$
In a wind shear, the tailwind suddenly decreases to just 10 kt. Due to the inertia of the aircraft, the ground speed cannot change instantaneously. Therefore, your TAS is now suddenly higher:
$$ \text{GS} = \text{TAS} + \text{TW} \; \Leftrightarrow \; \text{TAS} = \text{GS} - \text{TW} = 120 \, \text{kt} - 10 \, \text{kt} = 110 \, \text{kt} $$
If the aircraft was in trim, it will return to its trimmed airspeed eventually (assuming no pilot action). So this is just a short term in increase.


text:
Station pressure vs. reported altimeter
I must be missing something big.  Page 12-5 of the PHAK tells me that a station at 5,000 feet above sea level would convert its 24.92 pressure reading to 29.92.  I understand that this helps to track barometric pressure changes as it removes altitude from the equation.  But if that is the reported altimeter setting doesn’t it,well, defeat the whole purpose of taking feet above sea level into account for aviation purposes?  Obviously I’m missing something, but this is the way the PHAK words it.  Do they use another altimeter reading for aviation purposes, perhaps?  In other words they report a different setting than the one they converted to in the example above?


text:
Altimeter setting and pressure altitude
Student pilot here.  I thought I was smarter than this, I’m struggling with this concept.  I got the ASA test prep book and there’s a question asking what happens to the indicated altitude when the altimeter setting is changed from 29.15 to 29.85.  Answer is indicated altitude increases by 700 feet.  Ok, makes sense.
But then I’m reading the aircraft performance chapter in the Aeronautical Knowledge book.  It has a chart about finding pressure altitude, and when the altimeter setting decreases on this chart the altitude increases.  So, the converse of the concept in the question.  I can make either make sense on its own but they seem to contradict each other.  What am I missing?

I think your second paragraph is not as clear as it could be.  Which altitude increases?  Pressure altitude?  What is the reference on the chart?  Is it indicated altitude or actual altitude?  Do you have a reference?

Agree, I think you need to edit to " and when the altimeter setting decreases on this chart the PRESSURE altitude increases."  Right?

There’s a chart where you find the altimeter setting and it gives you the corresponding altitude correction.  Trying to post the pic but not sure how.  It has you add altitude to field elevation, as the altimeter setting decreases, to “get pressure altitude.”

@rbsc it looks like the answer solved your question, could you mark it as the answer so that other people can better find the solution when they have the same question?

Here's the difference in a nutshell:
When you change the number in Kollsman window to
bigger number and the atmosphere stays the same, the reading on your altimeter increases.
When you leave the number in the Kollsman window the same (let's say you have it set to 29.92 to see the pressure altitude) and the atmospheric conditions change so that the correct altimeter setting increases--i.e. the air pressure has increased-- then the reading on your altimeter decreases.
You increasing the setting in the Kollsman window is not the same thing as an increase in the prevailing correct altimeter setting.  They have the opposite effect on what you see on your altimeter dial.  The first thing makes the reading get higher, and the second thing makes the reading get lower.  When they both happen at the same time, the reading stays the same.
Clear as mud now?

Yes.  My brain was trying to tell me something like this but I couldn’t quite get there.  This absolutely clears it up, thanks so much.


text:
Why would payload optimized weight make a difference for bin space
I was on a Delta flight out of DTW and they told people (about 30 times at the gate and in the plane) that the flight is payload optimized and people needed to make sure they didn't put anything but a small carry on in the bin.
Why would this matter? The weight will be the same no matter where you put the luggage on the plane. Does vertical weight above the center of mass matter?

It's to prevent side-fumbling of the turbo encabulator. Sometimes, these silly officious announcements have far less basis in any technical necessity than in simply "hurry up, get in, sit down, and use the space at your feet before taking up limited space in the overhead bin," since if the latter runs out, the cabin crew has to gate-check any left-over bags, and that's extra work & extra time. "Payload optimized" sounds like techno babble to me - meant to sound impressive, but void of actual meaning.


text:
How do fighter jets stop pitching motion after a elevator deflection?
I'm working on a flight simulator , In which the jet has only ailerons , rudder , and two elvators at the back . When I nose up my aircraft via elevator deflection , it just keeps pitching up ( even when I let go of elevators and let them back to neutral)  , after going pitch max , it starts pitching in opposite direction and then back and forth . These oscillations take lot of time to die out and the motion  of my jet is like a sine wave going forward .
I'm using real forces of lift and weight on the aircraft body.
So my queries are ,

How does a fighter jet locks it's nose at a particular pitch angle ( the moment the pilot let go of the deflection) and  does it so smoothly and slowly , that it's possible to aim at whatever direction you like and hold it there , even when flying at 600KM/Hr.
Because when I let go of my elevators , entire plane just goes into oscillations and it becomes tedius task to focus on killing those oscillations .
A cheat that I'm thinking of is , let's say pilot makes +30 degree deflection to elevators and let go , then the computer will immediately put the elevators to -30 degreee deflection . This will exactly cancel out the rotational momentum built up by the initial deflection .  But in real aircraft , it will take time to put the control surface in opposite angle . So how do they do it so quickly .
This person claims to be using a PID controller for pitch damping . How does a PID controller acount for the inherent delay in the actuators of control surface . Let's say controller decides an +10 degree is needed this moment to keep the aircraft stable at the current pitch , but the hydraulics will take their own time to put the control surface at that angle , by that time aircraft will have moved even more from the required stable position .

I know I'm asking too much in a single question . But I've been stuck on this problem for so long that only more and more problems have accumulated in this time .

My guess to the first question: The pilot doesn't just 'let go'. When he wants to stop pitching up, he pushes the elevators down briefly to stop the rotation, and then returns the stick to centre to avoid inducing any other rotation.

"How does a PID controller acount for the inherent delay in the actuators of control surface" that's basic PID design and control theory, but answering here would take too long.

For me it sounds like your simulation is probably missing aerodynamic pitch dampening, which causes the plane to just keep rotating once a pitch moment was applied to the airframe.

Can you please provide a more detailed discription of your simulation

@hph304j Yes , right now each control surface can only deflect ,  exert lift force and drag force  NOTHING other than this . How do I go about implementing pitch damping that you speak bout . Any resource would help.

This depends on what level of realism you want.
If you want high, save yourself a lot of work by using an existing dynamics model like the JSBSim. There are model templates for it that have all the needed functions for reasonably realistic model are predefined and you just tweak the coefficients to get the performance you want.
If you want low, just make the surface deflection correspond to pitch/roll rate rather than acceleration, because that's what actually happens to a level of approximation that is perfectly fine for an arcade gamer.
What really happens is that when you deflect the control surface, the rotation will change the angle at which air hits the surface, and that will negate the effect of the control surface, so the aircraft settles at fixed rate of pitch/roll corresponding to the commanded displacement. And when the pilot centres the surface again, the aerodynamic force still exists that countered the deflection, which will now stop the rotation again.
This is fast enough that pilots normally don't need to stop the rotation with counter-movement of the controls except sometimes for fast rolls in aerobatics.
Regarding PID controller, a “position-integral-derivative” controller, the point is that the derivative term predicts the response delay, while the integral term compensates for any bias. The disadvantage is that if you tune it to have fast response, it always overshoots the target value, because the integral term will overestimate the bias. But it is simple to implement and does not require much modelling, just tweak the numbers until it works. The JSBSim provides PID controller (and some other simpler controllers) for the model definition.

There are two fundamental things that are (or should be) at play here: damping and natural pitch stability.
First, stability. Most airplanes (but not all fighters) have a natural tendency to come back to a certain (called trimmed) angle of attack. I won't explain here how it happens aerodynamically, but the point is that as the airplane pitches up, a moment arises that pitches it down (like a spring). This applies to any disturbance, be it a wind gust or an elevator deflection. So pulling (and holding) the elevator will not produce permanent rotation (unlike, say, the isolated roll response to ailerons). Instead, the airplane will settle on a higher angle of attack (AoA). (This will, of course, increase lift and the aircraft will depart from its former trajectory, but this is another matter).
Typically, you model this 'spring' by having a special function $C_m(\alpha)$ (pitch moment coefficient as a function of AoA (or of lift coefficient)). For a stable aircraft, it is linear and negative in the operational range of AoA. Which simply means: more AoA -> proportionally more pitch down moment. This is one of the most important functions in the whole flight simulation.
Elevator deflection (say, up) will add a pitch up moment. This will upset the original balance and will cause the airplane to accelerate pitching up. But this will only keep happening until the moment from the elevator balances out the growing pitch down moment from natural stability.
Now, the dynamics of this pitch process depends on several factors. Apart from this $C_m$ and the elevator deflection, it will depend on damping. Damping also happens naturally, and it happens for all airplanes, whether they are stable in pitch or not.
Damping, by definition, is the process that counteracts velocity: that is, the faster you move, the more the force that slows you down. In the absence of other forces, you will settle on some constant speed after initial acceleration. This, again, is something that would happen in an isolated roll motion: when the flow is symmetric, the wing doesn't produce any spring-like forces that restore bank.
There are two generic ways to simulate damping. One, you just directly introduce this term: $C_m(\dot\theta)$, i.e. pitch moment as a function of pitch rate. In the normal operational range of AoA, this is nearly a constant (i.e. a proportional term), and is always negative. The value of this constant is fairly easy to estimate from the tail geometry - see below. But things get more complicated as you approach stall. (They always do).
The second way is a simulation of how damping actually happens. Most of it comes from the fact that the local AoA on the tail changes as the aircraft pitches. Consider this: when you are pitching up, the tail goes down. More air "comes from below" to it, which is equivalent to higher AoA. The level of this increase depends on the added vertical speed of the tail with respect to the normal forward airspeed: $\arctan(w/v)$. (In practice you can usually drop arctan as the angle is small). Now, $w$ linearly depends on the pitch rate $\dot\theta$ (often denoted $q$) and the "arm" $L$ of the tail (its distance to CG): $q\cdot L$.
So, if you simulate the tail "properly", i.e. simulate lift on it and through that all the moments it produces, you'll need to add this local increase of its AoA, and everything will happen automatically. Alternatively, if you simulate the moments from the tail directly (which is often done), you can derive the amount of damping from knowing its nature: it (or rather its coefficient) will be proportional to $q\cdot L$, then to another $L$ because the force acts at this distance (making the dependency on $L^2$), and to the area of the horizontal stabiliser, and inversely to speed.
Tail is not the only contributor to damping, but usually it's by far the main one. If you account for it, your model will behave much better. Still, to account for all other things, you'll often have some additional term $C_{m_{other}}(q)$ (at least just linearly: $C_{m_{other}}^q\cdot q$), even if you simulate the tail separately.
This hopefully resolves the question why the model behaves so unnaturally. It remains to consider what happens if you indeed want to simulate a statically unstable fighter.
A statically unstable aircraft will still have the same damping, but its 'spring' term $C_m(\alpha)$ will be positive (or zero for neutral stability). If you pull the elevator up, the aircraft will start pitching up indefinitely. This motion will only be restrained in three ways:

Damping will limit the maximum pitch rate.
As the aircraft reaches stall (esp. tail stall), things will get very different. The aircraft may become stable again (leading to the Cobra-like manoeuvre), or may get "stuck" at certain high AoA. Simulating these regimes is a separate problem in flight dynamics.
Active computer control. Yes, a simple PID loop can solve it. Yes, delays in the control system is a real problem, and you will need more careful tuning of the controller, esp. its D term. It's fun.

I won't explain here how it happens aerodynamically That is perfectly fine, but even better would be to link to an answer that does.

@Peter, if you have a good candidate, you are welcome to edit it in (or suggest here).


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Why is Cathay Pacific flight path so long?
I was looking at flightradar24 and found this:

Which shows a flight from Vancouver, Canada to Hong Kong by Qatar Executive.
On the other hand I found this:

A flight from New York to Hong Kong by Cathay Pacific. But the flight path is so long. They are going to reach the North Pole first! Why is that flight path so long?
The direct links to the two flights:
Qatar Executive
Cathay Pacific

Ask yourself why the Qatar exec flight is going via Alaska instead of just a straight line across the Pacific. The answer to that will also answer the question about the Cathay flight.

Take a globe and pull a string between the two cities (departure and destination). That should give you the shortest route and answer this question.

@PeterKämpf But not all flight paths are great circles.

Possibly a dup of Why don't most planes fly in a straight path?, which explains why a circle appears as a complex path on Mercator maps.

@SnackExchange Most long-haul flights are reasonably close approximations to great-circles though. Deviation from such paths is usually a result of either safety concerns (see for example the number of commercial flights that avoid flyovers of Russia at the moment), sub-optimal prevailing winds (see for example flights from Norther Europe to the Eastern US, they go further north to avoid westerly winds) or airspace regulations.

A flight from Seoul, South Korea to Buenos Aires, Argentina could be done in two legs through any other intermediate city in the world without changing the total flight distance.

From Seoul to Buenos Aires is better to be Eastward. Isn't it?

Here is a picture of the direct path from New York to Hong Kong, as shown on a Globe, not a 2-D Map Projection.
The 2-D Map projection is very misleading.  The most direct route does involve heading north of Canada and Alaska.
Although, given the current conflict with Russia over Ukraine, I'm a little surprised the flight would pass over Russia at all.

The apparent length of these two flight paths is misleading due to the projection used on the map, which greatly increases the apparent size of paths at high latitude.
If you have a globe, try tracing the path you see on Flight Radar and comparing it to a path across the Pacific. You'll see the one that goes near the north pole is shorter.

In general, there may be other reasons not to choose the shortest path, leading to flight paths that seem overly long. For example, flights from the USA to the northern half of Europe are usually almost a straight line:

On the other hand, the return trips are usually much more north, like your Cathay Pacific flight:

The reason for this is are the winds: in the middle latitudes, they are usually westwards, while higher up North they are usually eastwards.


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Degree of motion terminology for special aviation movements
I perfectly understand yaw, pitch and roll. I'm wondering what a few special terms are for other unusual movements, such as those executed by a helicopter, drone or even perhaps something like a Harrier jump jet.

If "climb" refers to a straight up vertical motion along the z-axis, what is the name for the opposite? One might think fall, but I believe (keep me honest) in aviation, a "fall" is akin to a crash-landing type scenario, where the vehicle is out of control. Another word might be descent, but to me, that implies a downward change in pitch so that the vehicle gradually starts to lower its altitude over time and distance; instead I just want the opposite of a helicopter "climb"
I understand yaw means a change in left/right direction of the vehicle, but what motions describe a 90 degree, perpendicular change in motion to the left or right? Meaning, vehicle doesn't yaw, pitch or roll, it just "slides" 90 degrees to the left or right?

Any ideas as to what the correct terms are here?

For question 2 see What is the best word for lateral translational movement in a rotorcraft?

The opposite of climb is "descend" (that's an easy one).
The second one can be "translate" or possibly "slew" but I think the most appropriate word is "slip" as in sideslip.
Translate is a controlled movement from one place to another, and slew is an uncontrolled one according to Oxford, but when you work the "slew" switch for a directional gyro system, it's a controlled action, so there you go.
However, those terms apply to a stationary object that starts to move sideways, "slip" describes lateral non-turning movement of an airplane sideways when it's already moving forward, so this is probably the best definition.

"Descent" is the opposite of "climb".
"Translation" describes the second case.


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What classification of aerial vehicle would a manned quadcopter or octocopter be?
US here, if it makes a difference (w.r.t. FAA regulation, etc.).
There's a new upcoming class of aerial vehicles, similar to drones, however they are gigantic and they are manned -- that is, manned (piloted) quadcopters or octocopters like these ones.
I'm wondering what technical and/or legal classification(s) these types of vehicles would be considered to be. Any thoughts? Thanks in advance!

From 14 CFR 1.1:

Helicopter means a rotorcraft that, for its horizontal motion, depends principally on its engine-driven rotors.
Rotorcraft means a heavier-than-air aircraft that depends principally for its support in flight on the lift generated by one or more rotors.
Aircraft means a device that is used or intended to be used for flight in the air.

A manned quad- or octocopter seems to fit the definition for “helicopter”, assuming you mean the kind that have no fixed wing.

Wow, that's going to have to change.  That definition seems totally inappropriate--

Yeah, right? I don't see how those things require/necessitate the same type of regulation as a full bore helicopter....


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Using XFOIL to find the angle of attack corresponding to buffet onset and stall
I am trying to find the angle of attack that corresponds to the stall point and buffet onset. I understand for stall this occurs at the maximum lift coefficient? However upon plotting a range of angles on XFOIL it continues to ascend before it reaches a point where convergence no longer occurs so I am unsure what this means.

XFOIL uses potential code for purely subsonic flow. Per definition, it cannot be used for finding the AoA of buffet onset. Use ISES / MSES / MISES instead.

I may have used an incorrect term sorry- I mean like the boundary layer separation point for which the flow develops from laminar to transitional/turbulent if that makes a difference? Also, does this mean it cannot either be used to find the lift coefficient for which stall occurs? thanks

Boundary layer separation ≠ laminar to turbulent transition. Both are easy to find with XFOIL. Transition is part of regular output, separation can be found with the BL parameters output.


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Formation of Clouds in High Pressure Systems (Western Australia - Perth)
High pressure systems are normally correlated with stable atmospheric conditions with little to no cloud cover, but due to this minimal cloud cover, doesn't that open up the ocean / ground moisture to evaporate to then later condensate to form clouds?
I understand there are other forms of cloud formation like subsidence or frontal uplift, but some days here in Perth are very still and there hasn't been any fronts recently. For example during today (27/12/2021), it has been very still yet the heat has reached 40°C. For the previous two days there has also been temperatures reaching up to 45°C for prolonged periods of time with clear skies. I haven't felt much wind or any down drafts within my vicinity.
Today, there has been altostratus, altocumulus and cirrus clouds. Most of the clouds are isolated but some have formed extensive 'clumps' of cloud coverage with other clouds. So despite the high pressure system is it possible for there to be a decent amount of cloud generation? Can the ocean evaporate despite there being descending air?

I am not sure you can associate high pressure systems to high altitude cloud patterns. my understanding is that a high-pressure area is a region where the atmospheric pressure at the surface of the planet is greater than its surrounding environment.

High pressure means that there is descending air / 'the air above a given parcel of air is heavier than it'

But that has nothing to do with the formation of high altitude clouds.

If you read the answer, that is why I asked the question.

As always you could probably get a better answer on EarthScience.SE, but:
The centre of a high pressure system is less likely to see significant cloud yes, but the same is not true for the areas with closer isobars. For example, I was on the NSW southern coast that same day where we had miserably low temperatures and overcast low cloud. This is due to the same high pressure system affecting WA that was sitting south of the continent. In my case, because the wind flows anti-clockwise around a high in the southern hemisphere, this bad weather is essentially the ocean evaporation being blown onto us.
In Perth the story is a little different. The high is blowing hot desert air over your city, but the isobar spacing is reduced so therefore it has less force to it. There is not enough force to overcome the impact of that trough from the north. Whilst the trough officially ends slightly north of the city, it is still bringing significant moisture with it which can carry further south due to momentum. Furthermore, that isobar is actually the last associated with the high pressure system, and the gradient wind becomes an oceanic north-westerly just north of Perth.
The isobar also tells us the QNH around Perth is about 1012hPa which is neither high pressure nor low pressure. It is absolutely feasible in those conditions for there to be vertical uplift and instability to form clouds provided there is sufficient moisture - which the trough and the gentle north-westerly bring.

I already tried earth science, but there's been no people answering the question.


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Ground Speed vs True Airspeed from Departure Point to Destination - time interval
Looking at navigation theory right now and I'm stuck in the understanding of TAS and GS. How does GS determine the time interval it takes for an aircraft to fly from the DP to the DEST, if TAS determines how the fast the aircraft is really going through the air.
My current guess is that GS accounts for the wind, so that determines how fast it is going relative to the ground; thus determines the time interval.

Are you asking what the difference is between airspeed and groundspeed?

How does GS determine the time interval it takes for an aircraft to fly from the DP to the DEST, if TAS determines how the fast the aircraft is really going through the air.

The simple answer to this question is when you calculate the time to fly from DP to DEST you need to use distance and velocity t = d/v. But d and v need to be referenced to the same frame of reference.

DP and DEST locations are referenced to the ground frame.
The distance d is therefore computed in the ground frame.
v must be expressed in the ground frame, this is GS.

Computation of d in the ground frame
This is really simple but you need to know GS. If you know TAS and wind velocity, then the next method can be used.

While d and v must be in the same frame, this frame could be the air mass frame as well:

TAS is aircraft velocity measured relatively to the air mass frame.

To compute d in the air mass frame you would need to take into account that DP and DEST are moving relatively to the air mass frame due to the wind.
The distance d to cover is not the distance on the ground, but the distance between points in the air mass which coincide with the verticals of DP and DEST at the time the aircraft reaches these air mass points.

Computation of d in the air mass frame
Therefore to determine d you  would need to take into account the air mass velocity relative to the ground, that is the wind velocity (more accurately the wind velocity vector projected on the flight path).
This time you need to know both TAS and wind velocity.

My current guess is that GS accounts for the wind

As you infer correctly there is no mystery when using GS, GS is the sum of TAS and wind:

GS is related to TAS via wind velocity
So if you're lucky enough to know GS and use the ground frame, which is more simple, to compute the trip time, it's because someone or a tool already did the sum.

The third reference frame, the aircraft frame, could be used as well, d and v would need to be expressed in this frame. "I let the reader do the exercise" :-)


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Bernoulli Principle -- Reversed
The Bernoulli Principle states that as the velocity of a fluid increases, the pressure of that fluid decreases. However what if it were to be reversed: As the velocity of the fluid decreases, the pressure increases; is this possible?
I am looking at pressure differentials and how pressure is 'high' on the lower surface of the aerofoil. But is the pressure truly 'higher' or is it of normal pressure of the surroundings of the aerofoil?
So this is where the question comes in: If the pressure of the airflow is greater ont the lower surface of an aerofoil, then doesn't that mean that the airflow is slower? Or is the 'high pressure differential' just a word to help differenciate the concept of pressure on an aerofoil, and truly the pressure on the lower surface of the wing is normal?

This is an interesting hypothetical question, but I am not sure what it has to do with aviation?  Perhaps physics SE would be a better venue?

As said on Wikipedia, the principle says:
$$\frac{v^2}{2} + gz + \frac{p}{\rho} = \text{constant}$$
i.e., assuming no changes in the gravity potential energy or density:

an increase in velocity comes with a decrease in pressure
a decrease in velocity comes with an increase in pressure

It's all part of the same principle, there is no "reverse" needed.

So this is where the question comes in: If the pressure of the airflow
is greater ont the lower surface of an aerofoil, then doesn't that
mean that the airflow is slower?

Yes! But there is nothing technically 'reverse' in this. The Bernoulli Principle states that these parameters are linked, and if you increase one, the other decreases, and vice versa.
(Note that the pressure you are talking about is static pressure. The Principle effectively states that static pressure gets 'exchanged' for dynamic pressure.)

I am looking at pressure differentials and how pressure is 'high' on the lower surface of the aerofoil. But is the pressure truly 'higher' or is it of normal pressure of the surroundings of the aerofoil?

It is higher than at the top: that's how you get lift. But this is a very good thing to ponder. There is a very common misconception that the pressure below the wing is 'high'. This is not generally true. Oftentimes, esp. on airfoils with significant curvature of the lower surface (such as supercritical airfoils) the pressure on the bottom will be low, i.e. lower than ambient - but just 'less low' than above the wing.
Have a look at this question: What is the pressure on the pressure side in relation to the freestream? , and especially on this picture from it:

See this blue part? This is low pressure.


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Why don't hyperbolic navigation transmitters on the surface (e.g. Loran-C) allow altitude measurement?
I have been reading the following article about radio navigation in general and GPS / GLONASS receiver design:

A homemade receiver for GPS & GLONASS satellites

in which the author states the following about hyperbolic radionavigation:

Hyperbolic navigation systems were first implemented as ground-based navigation systems [...] Since the transmitters are located on the Earth's surface, the geometry of the problem does not allow a three-dimensional navigation. [...] To measure the altitude, one of the transmitters should be located above or below the user's receiver or at least out of the user's horizon plane.

I do not seem to be able to comprehend why this is the case. Even if the transmitters are located on the surface (e.g. Loran-C), the differences between reception times of signals from 4 transmitters would generate 3 rotational hyperboloids in 3D space, the intersection of which could be above (or below) surface.

You are correct, the time difference generates hyperboloids in 3D space and as such can be used to find the 3D position. However, if the transmitters are far apart and/or in the same plane, the geometry of the intersecting hyperboloids is such that the accuracy in the vertical axis is very poor.
When the receiver is in the same horizontal plane as the transmitters, all intersections almost form a vertical line. The slightest inaccuracy in time difference, would cause only a slight error in the horizontal position, but a significant error in the vertical position.
The concept is called Geometric Dilution of Precision (GDOP). A good start for more detail can be found on Wikipedia

The concept is called Geometric Dilution of Precision (GDOP). A good start for more detail can be found on Wikipedia.

Okay, I think I understand now. It's not that it's completely impossible to calculate the intersection with the transmitters on the same plane, it's that because the third hyperboloid is oriented the same as the other two (the lines connecting the receivers and transmitters are on the same plane), and given that the measurements are ambiguous, the "intersection volume" between the three ambiguous shapes would be high on the vertical axis, yes? (However that should not be the case for example if the third shape intersected the volume from an angle)

@rkourdis, that's right.

@Gerry, indeed. If I find the time I may expand on GDOP, HDOP and VDOP a bit.


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In F/A-18, where the servo-loop closure is accomplished?
In the flight control system of F/A-18, where is the servo-loop closed? I mean which of these two scenarios happen: 1-The position of the actuator and servo-valve ram are sent to the flight control computer and the servo-loop is closed there in the software, 2-the servo-loop is closed near actuator that means the actuator has an interface that receives a command and compares it with the actual position. If the second scenario is what happens, the interface is digital or analog?

Ram position feedback is sent back to the relevant channel in one of the two flight control computers via wire carrying an analogue signal. If it is a leading edge flap, aileron or rudder it is done twice. The big surfaces (Trailing Edge Flap and Horizontal Stab) are quad redundant. The signal is processed on a printed circuit board inside the box. There is negative feedback occurring via an operational amplifier type circuit.


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What do the Airway Letters and the following numbers mean?
For example, from RCTP - ZGGG the route is AJENT MKG A1 ANPOG R200 GLN W22 IDUMA via the AJENT SID to the IDUMA STAR. What does "A1 or R200 or W22" mean? Are they just randomly allocated letters with random numbers for routes?

RCTP - Taipei, Taiwan Airport
AJENT SID - Standard Instrument Departure leading to the Navigation fix named AJENT
MKG - fly to the MAGONG VOR Navigation beacon
A1 - follow airway A1
ANPOG - until the fix named ANPOG
R200 - follow airway R200
GLN - until the GUANLAN VOR Naviation beacon
W22 - follow airway W22
IDUMA - until the fix named IDUMA
IDUMA STAR - follow the IDUMA Standard Arrival Route
ZGGG - land at Guangzhou Baiyun International Airport

A1, R200 and W22 are airways. They are defined by lists of waypoints. The
waypoints between the first and last used on the airway are omitted and replaced by the airway name, otherwise the flightplan would be extremely long.
As an analogy one could describe a route on the road. For example Frankfurt - Amsterdam could be described as:
FRA - 44 - FRA KRZ - A3 - BORDER - A12 - KNP OUDENRIJN - A2 - KNP AMSTEL- S110 - AMS
Without mentioning all the highway intersections and exits, the route is described in sufficient detail to drive it. The numbers of the highways appear to be somewhat random, but I am sure somebody has thought about how to best assign the numbers.